数据结构_哈夫曼树及其应用

构造算法的例子

构造算法的实现

初始化，置权值

int i, m, s1, s2;
    m = 2 * n - 1;

    for (i = 1; i <= m; i++)
    {
        HT[i].lch = 0;
        HT[i].rch = 0;
        HT[i].parent = 0;
    }
    for (i = 1; i <= n; i++)
    {
        cin >> HT[i].weight;
    }

合并结点

// 创建哈夫曼树
    for (i = n + 1; i <= m; i++)
    {
        s1 = -1;
        s2 = -1;
        Selete(HT, i - 1, s1, s2);
        HT[s1].parent = i;
        HT[s2].parent = i;
        HT[i].lch = s1;
        HT[i].rch = s2;
        HT[i].weight = HT[s1].weight + HT[s2].weight;
    }

哈夫曼编码

void HuNode::create_Code(HuNode* HT, char** code, int n)
{
    int i, current, parent, k;
    char temp[100]; // 临时数组存放编码

    for (i = 1; i <= n; i++)
    {
        current = i;
        parent = HT[i].parent;
        k = 0; // 编码长度计数器

        while (parent != 0)
        {
            if (HT[parent].lch == current)
            {
                temp[k] = '0'; // 左子节点编码为 '0'
                k++;
            }
            else
            {
                temp[k] = '1';
                k++;
            }
            current = parent;
            parent = HT[current].parent;
        }
        temp[k] = '\0';

        // 将编码倒置并保存
        code[i] = new char[k + 1];
        for (int j = 0; j < k; j++)
        {
            code[i][j] = temp[k - j - 1];
        }
        code[i][k] = '\0';
    }
}

文件的编码或译码

int HuNode::Decode(const string codestr, char txtstr[], int n)
{
    int index, root, i, curNode;
    index = 0;
    root = 2 * n - 1; // 根节点编号
    curNode = root;

    for (i = 0; i < codestr.length(); i++)
    {
        if (codestr[i] == '0')
        {
            curNode = this[curNode].lch;
        }
        else
        {
            curNode = this[curNode].rch;
        }

        // 解码失败
        if (curNode == 0)
        {
            return error;
        }
        // 是叶子节点
        if (this[curNode].lch == 0 && this[curNode].rch == 0)
        {
            txtstr[index] = this[curNode].data;
            index++;
            curNode = root;
        }
    }

    if (curNode != root)
    {
        return error;
    }
    txtstr[index] = '\0';
    return ok;
}

原文地址：https://blog.csdn.net/2301_79837864/article/details/143577617

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